1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Time dependent perturbation for harmonic oscillator

  1. Mar 19, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm looking at the 1d harmonic oscillator
    \begin{equation}
    V(x)=\frac{1}{2}kx^2
    \end{equation}
    with eigenstates n and the time dependent perturbation
    \begin{equation}
    H'(t)=qx^3\frac{(\tau^2}{t^2+\tau^2}
    \end{equation}
    For t=-∞ the oscillator is in the groundstate n=0
    I need to show that for n>3 the states will not get excited and I only need to look at the first order perturbation.

    3. The attempt at a solution
    What I'm thinking is \begin{equation}P_{a→b}=\vert c_b(t)\vert^2 \end{equation} therefore I need to find when c_b=0
    and because
    \begin{equation}
    c_b'=-\frac{i}{\hbar}∫H_{ba}'(t')e^{i\omega_0t'}dt'
    \end{equation}
    I need to find
    \begin{equation}
    H_{ba}'=<\psi_n\vert H' \vert \psi_a >=0
    \end{equation}
    For n>3
    It would make sense if H_ba look a bit like this
    \begin{equation}
    H_{ba}'=\frac{d^n}{dx^n}H'
    \end{equation}
    But I get some horrible results if I try to solve H_ba. Is it right what I am doing or is it way off.
     
  2. jcsd
  3. Mar 19, 2012 #2

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You want to compute [itex]\langle n| H' |0\rangle[/itex], which is proportional to [itex]\langle n| \hat{x}^3 |0\rangle[/itex]. If you express [itex]\hat{x}[/itex] in terms of raising and lowering operators [itex]\hat{a},\hat{a}^\dagger[/itex], you should be able to see how to reach the necessary result.
     
  4. Mar 19, 2012 #3
    so if I use
    \begin{equation}
    \hat x=\sqrt{\frac{\hbar}{2m\omega}}(a_+ + a_-)
    \end{equation}

    then if I solved it right
    \begin{equation}
    \hat x^3 \psi_0=(\frac{\hbar}{2m\omega})^{\frac{3}{2}}[(a_+)^3+a_+ + a_- + (a_-)^3]\psi_0
    \end{equation}
    Is it correct if I use a_- on ψ_0 it will just be 0 so
    \begin{equation}
    \hat x^3 \psi_0=(\frac{\hbar}{2m\omega})^{\frac{3}{2}}[(a_+)^3\psi_0+a_+\psi_0]=(\frac{\hbar}{2m\omega})^{\frac{3}{2}}[\sqrt6 \psi_3+\psi_1]
    \end{equation}
    So it will end up like this
    \begin{equation}
    \langle \psi_n \vert\sqrt6 \psi_3+\psi_1 \rangle
    \end{equation}
    If this is correct I am not sure why the oscillator wouldn't get excited for n>3. Or is it my calculations there are wrong?

    *EDIT
    Oh is it because
    \begin{equation}
    \langle \psi_n \vert \psi_m \rangle = 0
    \end{equation}
    for n ≠ m
     
    Last edited: Mar 19, 2012
  5. Mar 19, 2012 #4

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You'll want to go back through the algebra here. Remember that [itex]a_+,a_-[/itex] don't commute, so you have to keep track of the order of terms.

    You will have to use both of these relations. You have all of the right ideas, just fix the algebra.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Time dependent perturbation for harmonic oscillator
Loading...